4 Linear Lie groups and their Lie algebras 4.1 Linear Lie groups 4.3 One-parameter subgroups

4.2 The exponential map

Recall that $\mathbb{K}$ denotes either $\mathbb{R}$ or $\mathbb{C}$ .

Definition 4.4.

Let $X\in\mathfrak{gl}_{n,\mathbb{K}}$ . We define

\exp(X)=\sum_{k=0}^{\infty}\frac{X^{k}}{k!}.

This series is convergent for all $X\in\mathfrak{gl}_{n,\mathbb{K}}$ . Let $||\cdot||$ be the matrix norm

||X||=\left(\sum_{i,j}|x_{ij}|^{2}\right)^{1/2}.

This satisfies the triangle inequality and also $||XY||\leq||X||||Y||$ — this can be proved using Cauchy–Schwarz. Then for any $X\in\mathfrak{gl}_{n,\mathbb{C}}$ with $||X||\leq M$ , we have

||\exp(X)||\leq\sum_{k=0}^{\infty}\frac{||X||^{k}}{k!}\leq\exp(M).

In particular, we see that $\exp$ is uniformly absolutely convergent on all compact subsets of $\mathfrak{gl}_{n,\mathbb{K}}$ . It follows that $\exp$ is a continuous function.

Lemma 4.5.

We have (for all $X,Y\in\mathfrak{gl}_{n,\mathbb{K}}$ , $s,t\in\mathbb{K}$ ):

1.

$\exp(0)=I$ .
2.

$\exp(X+Y)=\exp(X)\exp(Y)$ if $XY=YX$ . (This is NOT true in general).
3.

$\exp(X)$ is invertible, with inverse $\exp(-X)$ .
4.

$\exp(sX)\exp(tX)=\exp((s+t)X)$ .
5.

$g\exp(X)g^{-1}=\exp(gXg^{-1})$ .

Proof.

The first point is obvious. Let’s prove (2) from which (3) and (4) follow. By definition,

$\displaystyle\exp(X+Y)$	$\displaystyle=\sum_{k=0}^{\infty}\frac{(X+Y)^{k}}{k!}$
	$\displaystyle=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{\binom{k}{l}X^{l}Y^{k-l}}% {k!}$	(using that $X$ and $Y$ commute!)
	$\displaystyle=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{X^{l}Y^{k-l}}{l!(k-l)!}$
	$\displaystyle=\left(\sum_{l=0}^{\infty}\frac{X^{l}}{l!}\right)\left(\sum_{j=0}% ^{\infty}\frac{Y^{j}}{j!}\right),$	(putting $j=k-l$ )

which is equal to the right hand side. Rearranging the sums is valid by absolute convergence. Finally, (5) follows from $gX^{k}g^{-1}=(gXg^{-1})^{k}$ . ∎

In fact the exponential map is differentiable as a function of $X$ . For this, recall that a function $f:\mathbb{R}^{N}\rightarrow\mathbb{R}^{M}$ is differentiable at a point $p\in\mathbb{R}^{N}$ if there is a (necessarily unique) linear map $D_{p}f:\mathbb{R}^{N}\to R^{M}$ such that

\lim_{h\to 0}\frac{||f(p+h)-f(p)-D_{p}f(h)||}{||h||}=0,

and in this case $D_{p}f$ is called the derivative of $f$ at $p$ . (This definition is independent of the choice of norms on $\mathbb{R}^{N}$ and $\mathbb{R}^{M}$ ).

Proposition 4.6.

The exponential map is differentiable at the origin (zero matrix), and its derivative at the origin is the identity map from $\mathfrak{gl}_{n,\mathbb{C}}$ to itself.

Proof.

In the above definition we have, $N=M=2n^{2}$ , $f=\exp$ , $p=0$ , and we claim $D_{0}\exp$ is the identity. Thus we need to show

\lim\limits_{||X||\to 0}\frac{||\exp(X)-\exp(0)-X||}{||X||}=\lim\limits_{||X||% \to 0}\frac{||\exp(X)-I-X||}{||X||}=0,

which follows from the definition of the exponential map. Indeed,

\frac{||\exp(X)-I-X||}{||X||}=\frac{||\sum_{k=2}^{\infty}\frac{X^{k}}{k!}||}{|% |X||}\leq||X||\cdot\sum_{k=0}^{\infty}\frac{||X||^{k}}{(k+2)!}<||X||e^{||X||},

which tends to zero as $||X||\to 0$ . ∎

Remark 4.7.

In fact, the exponential function has derivatives to all orders at all points; this follows from the fact that it is given by power series that converge absolutely at all points and all of whose (formal) derivatives also converge absolutely at all points.

By the inverse function theorem, it follows from the remark that

Corollary 4.8.

The exponential map is a local diffeomorphism at $0$ : there exist neighbourhoods $U_{0}\subset\mathfrak{gl}_{n,\mathbb{K}}$ containing $0$ and $V_{0}\subset\operatorname{GL}_{n}(\mathbb{K})$ containing $I$ such that $\exp|_{U_{0}}$ is a smooth homeomorphism onto $V_{0}$ with smooth inverse.

Remark 4.9.

In face we can take $V_{0}=\{X\in\operatorname{GL}_{n}(\mathbb{C}):||X-I||<1\}$ . The inverse of $\exp$ in this neighbourhood is

\log(X)=\sum_{k=0}^{\infty}(-1)^{k}\frac{(X-I)^{k+1}}{k+1},

which is convergent when $||X-I||<1$ .

Of course, $\exp$ is not injective in general. For example, $\exp(2\pi ik)=1$ for $k\in\mathbb{Z}$ .

For the next result it will be useful to know the following facts from linear algebra.

Lemma 4.10.

Let $X\in\operatorname{GL}_{n}(\mathbb{C})$ . Then $X$ is conjugate to a matrix of the form $D U$ where

•

$D$ is diagonal
•

$U$ is upper triangular with ‘1’s on the diagonal
•

$D$ and $U$ commute.

Proof.

(nonexaminable) This follows from Jordan normal form. Here’s a direct proof. Firstly write $\mathbb{C}^{n}$ as a direct sum of generalised eigenspaces for $X$ : if $\lambda$ is an eigenvalue of $X$ then we can write the characteristic polynomial $P(T)=(T-\lambda)^{a}Q(T)$ where $Q(T)$ does not have $\lambda$ as a root and $a\geq 1$ is an integer. Then the image of $Q(X)$ on $\mathbb{C}^{n}$ is the generalised eigenspace of $\lambda$ . The kernel of $Q(X)$ is preserved by $X$ and $X$ does not have an eigenvalue equal to $\lambda$ since $\lambda$ is not a root of $Q(X)$ , which must be the characteristic polynomial of $X$ acting on $\ker Q(X)$ . Thus

\operatorname{im}Q(X)\cap\ker Q(X)=\{0\}

and by the rank-nullity theorem

\mathbb{C}^{n}=\operatorname{im}Q(X)\oplus\ker Q(X)

is a decomposition of $\mathbb{C}^{n}$ as a direct sum of the $\lambda$ generalised eigenspace and a subspace preserved by $X$ . Repeating for each eigenvector gives the required decomposition of $\mathbb{C}^{n}$ . This reduces the proof of the statement to the case where $X$ has only one eigenvalue $\lambda$ . In this case, we can inductively choose a basis $v_{1},\ldots,v_{n}$ of $\mathbb{C}^{n}$ such that, for $1\leq i\leq n$ , the image of $v_{i}$ in $\mathbb{C}/\left\langle v_{1},\ldots,v_{i-1}\right\rangle$ is an eigenvector of $X$ with eigenvalue $\lambda$ . With respect to this basis, $X$ is then diagonal with $\lambda$ ’s on the diagonal, and we get the required decomposition with $D=\lambda I$ . ∎

Lemma 4.11.

The exponential function $\exp:\mathfrak{gl}_{n,\mathbb{C}}\to\operatorname{GL}_{n}(\mathbb{C})$ is surjective.

Proof.

First prove it for $D$ and $U$ as in Lemma 4.10. The case of diagonal matrices is easy (homework!) whereas for $U$ you can use that the power series for $\log(U)$ in terms of powers of $U-I$ is actually a polynomial (homework!).

For general $X$ , by conjugating (homework!) we can reduce to the case where $X=DU$ as above. If $D=\exp(d)$ and $U=\exp(u)$ then

DU=\exp(d)\exp(u)=\exp(d+u)

because $d$ and $u$ commute (so long as you choose $d$ and $u$ carefully — note that $\exp$ isn’t injective — homework!). ∎

Remark 4.12.

The lemma is not true over $\mathbb{R}$ ; as we will see, the determinant of $\exp(X)$ is positive for all real matrices $X$ .

Lemma 4.13.

We have

\det\exp(X)=\exp\operatorname{tr}(X).

Proof.

Conjugate so that $X$ is an upper triangular matrix with diagonal entries $\lambda_{1},\ldots,\lambda_{n}$ , and then note that $\exp(X)$ is also diagonal with entries $\exp(\lambda_{1}),\ldots,\exp(\lambda_{n})$ (in the lecture I overcomplicated this!).

Thus

\det\exp(X)=\prod_{i=1}^{n}\exp(\lambda_{i})=\exp\left(\sum_{i=1}^{n}\lambda_{% i}\right)=\exp\operatorname{tr}(X).

∎

The next proposition will be useful when we discuss Lie algebras of linear Lie groups.

Proposition 4.14.

(Lie product formula) We have

\exp(X+Y)=\lim_{k\to\infty}\left(\exp\left(\frac{X}{k}\right)\exp\left(\frac{Y% }{k}\right)\right)^{k}.

Proof.

Note that $\exp(tX)\exp(tY)=\exp(t(X+Y)+O(t^{2}))$ as $t\to 0$ , as follows by taking $\log$ of the left hand side. Therefore

	$\displaystyle\left(\exp\left(\frac{X}{k}\right)\exp\left(\frac{Y}{k}\right)% \right)^{k}$	$\displaystyle=\left(\exp\left(\frac{X+Y}{k}+O\left(\frac{1}{k^{2}}\right)% \right)\right)^{k}$
		$\displaystyle=\exp\left(X+Y+O\left(\frac{1}{k}\right)\right).$

Now take the limit as $k\to\infty$ . ∎